2 Answers
2

Let $N$ be a set of measure zero. Then for every $\varepsilon \gt 0$ we can write $N \subset \bigcup_{k=1}^\infty B_k$ where each $B_k$ is a ball of radius $r_k$ and $\sum_k \mu B_k \lt \varepsilon$. But then by Lipschitz continuity $f(B_k)$ is contained in a ball of radius $L\cdot r_k$ where $L$ is the Lipschitz constant of $f$, and thus $\mu^\ast f(B_k) \leq L^d \mu B_k$ so that
$$
\mu^\ast f(N) \leq L^d \sum\nolimits_k \mu B_k \lt L^d \varepsilon.
$$
As $\varepsilon \gt 0$ was arbitrary, we see that $f(N)$ must have zero outer measure and hence it is a null set.

so let $A \subseteq \mathbb R^d$ be a set with measure zero. For $n \in \mathbb N$ let $A_n := B_n(0) \cap A$. Then $A_n$ has measure zero as $A_n \subseteq A$ for each $n$ and we have $A = \bigcup_n A_n$. From your bounded case you get that $f(A_n)$ is of measure zero for each $n$. But as
\[
f(A) = f\left(\bigcup_n A_n\right) = \bigcup_n f(A_n)
\]
$f(A)$ is a countable union of sets of measure zero and has therefore measure zero.